CHAPTER 5:Chemical Kinetics
Copyright © 2020 by Nob Hill Publishing, LLC
The purpose of this chapter is to provide a framework for determining the reaction rate given a detailed statement of the reaction chemistry. We use several concepts from the subject of chemical kinetics to illustrate two key points: 1 The stoichiometry of an elementary reaction defines the concentration dependence of the rate expression. 2 The quasi-steady-state assumption (QSSA) and the reaction equilibrium assumption allow us to generate reaction-rate expressions that capture the details of the reaction chemistry with a minimum number of rate constants.
1 / 152 The Concepts
The concepts include: the elementary reaction Tolman’s principle of microscopic reversibility elementary reaction kinetics the quasi-steady-state assumption the reaction equilibrium assumption The goals here are to develop a chemical kinetics basis for the empirical expression, and to show that kinetic analysis can be used to take mechanistic insight and describe reaction rates from first principles.
2 / 152 Reactions at Surfaces
We also discuss heterogeneous catalytic adsorption and reaction kinetics. Catalysis has a significant impact on the United States economy1 and many important reactions employ catalysts.2 We demonstrate how the concepts for homogeneous reactions apply to heterogeneously catalyzed reactions with the added constraint of surface-site conservation. The physical characteristics of catalysts are discussed in Chapter 7.
1Catalysis is at the heart of the chemical industry, with sales in 1990 of $292 billion and employment of 1.1 million, and the petroleum refining industry, which had sales of $140 billion and employment of 0.75 million in 1990 [13]. 2Between 1930 and the 1980s, 63 major products and 34 major process innovations were introduced by the chemical industry. More than 60% of the products and 90% of these processes were based on catalysis[13]. 3 / 152 Elementary Reactions and Microscopic Reversibility
Stoichiometric statements such as
A + B )−*− C
are used to represent the changes that occur during a chemical reaction. These statements can be interpreted in two ways. The reaction statement may represent the change in the relative amounts of species that is observed when the reaction proceeds. overall stoichiometry Or the reaction statement may represent the actual molecular events that are presumed to occur as the reaction proceeds. elementary reaction The elementary reaction is characterized by a change from reactants to products that proceeds without identifiable intermediate species forming.
4 / 152 Elementary Reactions and Rate Expressions
For an elementary reaction, the reaction rates for the forward and reverse paths are proportional to the concentration of species taking part in the reaction raised to the absolute value of their stoichiometric coefficients. The reaction order in all species is determined directly from the stoichiometry. Elementary reactions are usually unimolecular or bimolecular because the probability of collision between several species is low and is not observed at appreciable rates. For an overall stoichiometry, on the other hand, any correspondence between the stoichiometric coefficients and the reaction order is purely coincidental.
5 / 152 Some Examples — Acetone decomposition
The first example involves the mechanism proposed for the thermal decomposition of acetone at 900 K to ketene and methyl-ethyl ketone (2-butanone) [17]. The overall reaction can be represented by
3CH3COCH3 −→ CO + 2CH4 + CH2CO + CH3COC2H5
6 / 152 Acetone decomposition
The reaction is proposed to proceed by the following elementary reactions
3CH3COCH3 −→ CO + 2CH4 + CH2CO + CH3COC2H5
CH3COCH3 −→ CH3 + CH3CO
CH3CO −→ CH3 + CO
CH3 + CH3COCH3 −→ CH4 + CH3COCH2
CH3COCH2 −→ CH2CO + CH3
CH3 + CH3COCH2 −→ CH3COC2H5
7 / 152 Acetone decomposition
The thermal decomposition of acetone generates four stable molecules that can be removed from the reaction vessel: CO, CH4, CH2CO and CH3COC2H5. Three radicals are formed and consumed during the thermal decomposition of acetone: CH3, CH3CO and CH3COCH2. These three radicals are reaction intermediates and cannot be isolated outside of the reaction vessel.
8 / 152 Overall stoichiometry from Mechanism
Assign the species to the A vector as follows Species Formula Name A1 CH3COCH3 acetone A2 CH3 methyl radical A3 CH3CO acetyl radical A4 CO carbon monoxide A5 CH3COCH2 acetone radical A6 CH2CO ketene A7 CH4 methane A8 CH3COC2H5 methyl ethyl ketone The stoichiometric matrix is −1 1 1 0 0 0 0 0 0 1 −1 1 0 0 0 0 ν = −1 −1 0 0 1 0 1 0 0 1 0 0 −1 1 0 0 0 −1 0 0 −1 0 0 1
9 / 152 Overall stoichiometry from Mechanism
Multiply ν by 1 1 2 1 1 We obtain −3 0 0 1 0 1 2 1 which is the overall stoichiometry given in Reaction 5.1.
10 / 152 Example Two — Smog
The second example involves one of the major reactions responsible for the production of photochemical smog. The overall reaction and one possible mechanism are
2NO2 + hν −→ 2NO + O2
NO2 + hν −→ NO + O
O + NO2 )−*− NO3
NO3 + NO2 −→ NO + O2 + NO2
NO3 + NO −→ 2NO2
O + NO2 −→ NO + O2
11 / 152 Smog
In this reaction mechanism, nitrogen dioxide is activated by absorbing photons and decomposes to nitric oxide and oxygen radicals (elementary Reaction 5.8).
Stable molecules are formed: NO and O2
Radicals, O and NO3, are generated and consumed during the photochemical reaction. Work through the steps to determine the linear combination of the mechanistic steps that produces the overall reaction.
12 / 152 Example Three — Methane Synthesis
The third example involves the synthesis of methane from synthesis gas, CO and H2, over a ruthenium catalyst [6]. The overall reaction and one possible mechanism are
3H2(g) + CO(g) −→ CH4(g) + H2O(g)
CO(g) + S )−*− COs
COs + S )−*− Cs + Os
Os + H2(g) −→ H2O(g) + S
H2(g) + 2S )−*− 2Hs
Cs + Hs )−*− CHs + S
CHs + Hs )−*− CH2s + S
CH2s + Hs )−*− CH3s + S
CH3s + Hs −→ CH4(g) + 2S
13 / 152 Methane Synthesis
The subscripts g and ads refer to gas phase and adsorbed species, respectively, and S refers to a vacant ruthenium surface site. During the overall reaction, the reagents adsorb (elementary Reactions 5.150 and 5.153), and the products form at the surface and desorb (elementary Reactions 5.152 and 5.157).
Adsorbed CO (COs) either occupies surface sites or dissociates to adsorbed carbon and oxygen in elementary Reaction 5.151.
14 / 152 Methane Synthesis
The adsorbed carbon undergoes a sequential hydrogenation to methyne, methylene, and methyl, all of which are adsorbed on the surface. Hydrogenation of methyl leads to methane in elementary Reaction 5.157. In this example, the overall reaction is twice the fourth reaction added to the remaining reactions.
15 / 152 Elementary Reactions — Characteristics
One criterion for a reaction to be elementary is that as the reactants transform into the products they do so without forming intermediate species that are chemically identifiable. A second aspect of an elementary reaction is that the reverse reaction also must be possible on energy, symmetry and steric bases, using only the products of the elementary reaction. This reversible nature of elementary reactions is the essence of Tolman’s principle of microscopic reversibility [19, p. 699].
16 / 152 Microscopic Reversibility
This assumption (at equilibrium the number of molecules going in unit time from state 1 to state 2 is equal to the number going in the reverse direction) should be recognized as a distinct postulate and might be called the principle of microscopic reversibility. In the case of a system in thermodynamic equilibrium, the principle requires not only that the total number of molecules leaving a given quantum state in unit time is equal to the number arriving in that state in unit time, but also that the number leaving by any one particular path, is equal to the number arriving by the reverse of that particular path.
17 / 152 Transition State Theory
Various models or theories have been postulated to describe the rate of an elementary reaction. Transition state theory (TST)is reviewed briefly in the next section to describe the flow of systems (reactants −→ products) over potential-energy surfaces. Using Tolman’s principle, the most probable path in one direction must be the most probable path in the opposite direction. Furthermore, the Gibbs energy difference in the two directions must be equal to the overall Gibbs energy difference — the ratio of rate constants for an elementary reaction must be equal to the equilibrium constant for the elementary reaction.
18 / 152 Elementary Reaction Kinetics
In this section we outline the transition-state theory (TST), which can be used to predict the rate of an elementary reaction. Our purpose is to show how the rate of an elementary reaction is related to the concentration of reactants. The result is
Y −νij Y νij ri = ki cj − k−i cj (5.22) j∈Ri j∈Pi in which j ∈ Ri and j ∈ Pi represent the reactant species and the product species in the ith reaction, respectively. Equation 5.22 forms the basis for predicting reaction rates and is applied to homogeneous and heterogeneous systems. Because of its wide use, much of Chapter 5 describes the concepts and assumptions that underlie Equation 5.22.
19 / 152 Potential Energy Diagrams
We consider a simple reaction to illustrate the ideas
F + H H )−*− {F H H} )−*− F H + H
The potential energy V (r) of the molecule is related to the relative position of the atoms, and for a diatomic molecule the potential energy can be represented with a Morse function h i V (r) = D e−2βr − 2e−βr in which D is the dissociation energy, r is the displacement from the equilibrium bond length, and β is related to the vibrational frequency of the molecule. The following figure presents the Morse functions for HF and H2 molecules.
20 / 152 Potential energy of HF and H2 molecules
200
150 HHF
100 rHF rHH
50
0 V (kcal/mol)
−50
HH −100 HF
−150 0 1 2 3 4 5
rHH, rHF (A)˚
Figure 5.1: Morse potential for H2 and HF.
21 / 152 Coordinate system for multi-atom molecules
Molecules containing N atoms require 3N Cartesian coordinates to specify the locations of all the nuclei. Consider a moleculre of three atoms; we require only three coordinates to describe the internuclear distances. We choose the two distances from the center atom and the bond angle as the three coordinates. The potential energy is a function of all three coordinate values. A plot of the energy would be a three-dimensional surface, requiring four dimensions to plot the surface.
22 / 152 Coordinate system
We fix the bond angle by choosing a collinear molecule Only the two distances relative to a central atom are required to describe the molecule. The potential energy can be expressed as a function of these two distances, and we can view this case as a three-dimensional plot.
23 / 152 Collinear Approach of HF and H2
Figure 5.2 shows a representative view of such a surface for the collinear collision between F and H2.
F + H H )−*− {F H H} )−*− F H + H
Several sources provided the data used to calculate this potential-energy surface [14, 16, 15]. Moore and Pearson provide a more detailed discussion of computing potential-energy surfaces [12].
24 / 152 Collinear Approach of HF and H2
V (kcal/mol) 0 −40 −80 −120 0 0.5 1 0 0.5 1.5 r (A)˚ 1 HH 2 1.5 2 2.5 rHF (A)˚ 2.5 3 3
Figure 5.2: Potential-energy surface for the F, H2 reaction.
25 / 152 Collinear Approach of HF and H2 – Energy Contours
3
2.5
2 ˚ A) ( 1.5 HH r
1
0.5
0 0 0.5 1 1.5 2 2.5 3
rHF (A)˚
Figure 5.3: Contour representation of the potential-energy surface.
26 / 152 The energy surface
Large rHF. F atoms exist along with H2 molecules. Reproduces the H2 curve in Figure 5.1.
Large rHH. H atoms exist along with HF molecules. Reproduces the curve for HF in Figure 5.1.
As the F atom is brought into contact with H2 in a collinear fashion, a surface of energies is possible that depends on the values of rHF and rHH. The minimum energy path along the valleys is shown by the dashed line in Figures 5.2 and 5.3.
The reaction starts in the H2 valley at large rHF, proceeds along the minimum energy path and ends in the HF valley at large rHH.
27 / 152 The energy surface
The transistion state is is the saddle point connecting the two vallyes. the minimum energy path is a saddle point. The dashed line is referred to as the reaction-coordinate pathway, and movement along this path is the reaction coordinate, ξ. Figure 5.4 is a reaction-coordinate diagram, which displays the energy change during this reaction.
28 / 152 Reaction coordinate diagram
-90
-95
-100
-105
-110 (kcal/mol)
V -115
-120
-125
-130 0 1 2 3 4 5 6 7 ξ
Figure 5.4: Reaction-coordinate diagram.
29 / 152 Reaction coordinate diagram
The reaction coordinate represents travel along the minimum energy path from reactants to products. The difference in energies between reactants and products is the heat of reaction; in this case it is exothermic by 34 kcal/mol. The barrier height of 4 kcal/mol between reactants and products is used to calculate the activation energy for the reaction.
30 / 152 Transition state complex
The molecular structure (the relative positions of F, and two H’s) at the saddle point is called the transition-state complex. The principle of microscopic reversibility implies that the same structure at the saddle point must be realized if the reaction started at HF + H and went in reverse. The transition-state complex is NOT a chemically identifiable reaction intermediate; it is the arrangement of atoms at the point in energy space between reactants and products. Any change of relative positions of F and the two H’s for the transition-state complex (as long as the motions are collinear) results in the complex reverting to reactants or proceeding to products. Using statistical mechanics and the potential energy diagram, we can calculate the rate constant, i.e. the frequency at which the transitions state decomposes to reactants or products.
31 / 152 Fast and Slow Time Scales
One of the characteristic features of many chemically reacting systems is the widely disparate time scales at which reactions occur. It is not unusual for complex reaction mechanisms to contain rate constants that differ from each other by several orders of magnitude. Moreover, the concentrations of highly reactive intermediates may differ by orders of magnitude from the concentrations of relatively stable reactants and products. These widely different time and concentration scales present challenges for accurate estimation of rate constants, measurement of low-concentration species, and even numerical solution of complex models.
32 / 152 Fast and Slow Time Scales
On the other hand, these disparate scales often allow us to approximate the complete mechanistic description with simpler rate expressions that retain the essential features of the full problem on the time scale or in the concentration range of interest. Although these approximations were often used in earlier days to allow easier model solution, that is not their primary purpose today. The reduction of complex mechanisms removes from consideration many parameters that would be difficult to estimate from available data. The next two sections describe two of the most widely used methods of model simplification: the equilibrium assumption and the quasi-steady-state assumption.
33 / 152 The Reaction Equilibrium Assumption
In the reaction equilibrium assumption, we reduce the full mechanism on the basis of fast and slow reactions. In a given mechanism consisting of multiple reactions, some reactions may be so much faster than others, that they equilibrate after any displacement from their equilibrium condition. The remaining, slower reactions then govern the rate at which the amounts of reactants and products change. If we take the extreme case in which all reactions except one are assumed at equilibrium, this remaining slow reaction is called the rate-limiting step. But the equilibrium approximation is more general and flexible than this one case. We may decompose a full set of reactions into two sets consisting of any number of slow and fast reactions, and make the equilibrium assumption on the set of fast reactions.
34 / 152 The archetype example of fast and slow reactions
We illustrate the main features with the simple series reaction
k1 k2 A )−*− B, B )−*− C k−1 k−2
Assume that the rate constants k2, k−2 are much larger than the rate constants k1, k−1, so the second reaction equilibrates quickly. By contrast, the first reaction is slow and can remain far from equilibrium for a significant period of time. Because the only other reaction in the network is at equilibrium, the slow, first reaction is called the rate-limiting step. We show that the rate of this slow reaction does indeed determine or limit the rate at which the concentrations change.
35 / 152 The complete case
We start with the full model and demonstrate what happens if k2, k−2 k1, k−1. Consider the solution to the full model with rate constants k1 = 1, k−1 = 0.5, k2 = k−2 = 1 and initial conditions cA(0) = 1.0, cB (0) = 0.4, cC (0) = 0 shown in Figure 5.6.
36 / 152 Full model, rates about the same
1
0.8 cA
0.6 cB
0.4 concentration
0.2 cC
0 0 1 2 3 4 5 t
Figure 5.6: Full model solution for k1 = 1, k−1 = 0.5, k2 = k−2 = 1.
37 / 152 Turn up the second rate
1
0.8 cA
0.6
0.4 cB concentration
0.2 cC
0 0 1 2 3 4 5 t
Figure 5.7: Full model solution for k1 = 1, k−1 = 0.5, k2 = k−2 = 10.
38 / 152 Turn it way up
0.4
0.35
cB 0.3
0.25
0.2
concentration 0.15
0.1 cC k2 = 10 k = 20 0.05 2 k2 = 50 k2 = 100 0 0 0.05 0.1 0.15 0.2 0.25 t
Figure 5.8: Concentrations of B and C versus time for full model with increasing k2 with K2 = k2/k−2 = 1.
39 / 152 Now let’s do some analysis
We now analyze the kinetic model. The text shows how to derive models for both the fast and slow time scales. The text also treats models in terms of reaction extent and concentration variables. Here I will focus on only the slow time scale model, and only the concentration variables. The main thing we are trying to do is enforce the condition that the second reaction is at equilibrium. We therefore set its rate to zero to find the equilibrium relationship between cB and cC .
40 / 152 Equilibrium constraint
The equilibrium assumption is made by adding the algebraic constraint that r2 = 0 or K2cB − cC = 0.
r2 = 0 = k2cB − k−2cC
k2 0 = K2cB − cC , K2 = k−2 The mole balances for the full model are dc dc dc A = −r , B = r − r , C = r dt 1 dt 1 2 dt 2
We can add the mole balances for components B and C to eliminate r2 d(c + c ) B C = r (5.45) dt 1
Notice we have not set r2 = 0 to obtain this result. Equation 5.45 involves no approximation.
41 / 152 Getting back to ODEs
If we differentiate the equilibrium condition, we obtain a second relation, K2dcB /dt − dcC /dt = 0, which allows us to solve for both dcB /dt and dcC /dt,
dcB 1 dcC K2 = r1 = r1 (∗) dt 1 + K2 dt 1 + K2
42 / 152 Equations Fast Time Scale(τ = k−2t) Slow Time Scale(t)
dcA dcA = 0, c (0) = c = −r1, c (0) = c dτ A A0 dt A A0 dcB dcB 1 ODEs = −(K2cB − cC ), cB (0) = cB0 = r1, cB (0) = cBs dτ dt 1+K2
dcC dcC K2 = K2cB − cC , cC (0) = cC0 = r1, cC (0) = cCs dτ dt 1+K2 dcA c = c = −r1, c (0) = c A A0 dt A A0 dcB DAEs = −(K2c − c ), c (0) = c 0 = K2c − c dτ B C B B0 B C
0 = cA − cA0 + cB − cB0 + cC − cC0 0 = cA − cA0 + cB − cB0 + cC − cC0
1 K2 cBs = (cB0 + cC0), cCs = (cB0 + cC0) 1+K2 1+K2
43 / 152 Full model and equilibrium assumption
1 full model equilibrium assumption
0.8
cA
0.6
0.4 cB concentration
0.2 cC
0 0 0.5 1 1.5 2 2.5 3 t
Figure 5.9: Comparison of equilibrium assumption to full model for k1 = 1, k−1 = 0.5, k2 = k−2 = 10.
44 / 152 A trap to avoid
Critique your colleague’s proposal. “Look, you want to enforce cC = K2cB , right? Well, that means r2 = 0, so just set it to zero, and be done with it. I get the following result.”
dc dc dc A = −r B = r − r C = r full model dt 1 dt 1 2 dt 2 dc dc dc A = −r B = r C = 0 reduced model dt 1 dt 1 dt The result for B and C doesn’t agree with Equation (∗). What went wrong?
45 / 152 Paradox lost
Let’s look at C’s mass balance again dc C = k c − k c dt 2 B −2 C
= k−2(K2cB − cC )
Your colleague is suggesting this product is zero because the second term is zero, but what did we do to make the full model approach equilibrium?
We made k2, k−2 large!
46 / 152 Paradox lost
So here’s the flaw. dcC = k−2 (K2cB − cC ) 6= 0 dt |{z} | {z } goes to ∞! goes to 0 The product does not go to zero. In fact we just showed
dcC K2 = r1 dt 1 + K2 This fact is far from obvious without the analysis that we did.
47 / 152 Further special cases – second reaction irreversible
We start with the slow time-scale model given in the table and take the limit as K2 −→ ∞ giving
dc A = −r , c (0) = c dt 1 A A0 dc B = 0, c (0) = 0 dt B dc C = r , c (0) = c + c dt 1 C B0 C0
In this limit the B disappears, cB (t) = 0, and r1 = k1cA, which describes the irreversible reaction of A directly to C with rate constant k1
k A −→1 C
We see in the next section that this case is well described also by making the quasi-steady-state assumption on species B.
48 / 152 Second reaction irreversible in other direction
Take the second reaction as irreversible in the backward direction, K2 = 0. We obtain from the table dc A = −r , c (0) = c dt 1 A A0 dc B = r , c (0) = c + c dt 1 B B0 C0 dc C = 0, c (0) = 0 dt C This describes the reversible reaction between A and B with no second reaction, cC (t) = 0,
k1 A )−*− B k−1 Under the equilibrium assumption, we see that the two series reactions in Equation 5.41 may reduce to A going directly and irreversibly to C, Equation 5.46, or A going reversibly to B, Equation 5.47, depending on the magnitude of the equilibrium constant of the second, fast reaction.
49 / 152 The Quasi-Steady-State Assumption
In the quasi-steady-state assumption, we reduce the full mechanism on the basis of rapidly equilibrating species rather than reactions as in the reaction equilibrium assumption. Highly reactive intermediate species are usually present only in small concentrations. Examples of these transitory species include atoms, radicals, ions and molecules in excited states.
50 / 152 The Quasi-Steady-State Assumption
We set the production rate of these species equal to zero, which enables their concentration to be determined in terms of reactants and products using algebraic equations. After solving for these small concentration species, one is then able to construct reaction-rate expressions for the stable reactants and stable products in terms of reactant and product concentrations only. The idea to set the production rate of a reaction intermediate equal to zero has its origins in the early 1900s [2, 4] and has been termed the Bodenstein-steady-state, pseudo-steady-state, or quasi-steady-state assumption. The intermediate species are referred to as QSSA species [20].
51 / 152 QSSA Example
To illustrate the QSSA consider the simple elementary reaction
k k A −→1 BB −→2 C
Let the initial concentration in a constant-volume, isothermal, batch reactor be cA = cA0, cB = cC = 0. The exact solution to the set of reactions is
−k1t cA = cA0e
k1 −k1t −k2t cB = cA0 e − e k2 − k1
1 −k1t −k2t cC = cA0 k2(1 − e ) − k1(1 − e ) (5.52) k2 − k1
52 / 152 Making B a highly reactive intermediate species
We make k2 >> k1 in order for B to be in low concentration. Formation reactions are much slower than consumption reactions. Figure 5.10 shows the normalized concentration of C given in Equation 5.52 versus time for different values of k2/k1.
−k1t The curve cC /cA0 = 1 − e is shown also (the condition where k2/k1 = ∞).
Figure 5.10 illustrates that as k2 k1, the exact solution is equivalent to
−k1t cC = cA0 1 − e (5.53)
53 / 152 Full model
1
0.9
0.8
0.7
0.6
cC 0.5 cA0 0.4
0.3
0.2 k2/k1 = 1 k2/k1 = 10 0.1 k2/k1 = 50 k2/k1 = ∞ 0 0 1 2 3 4 5
k1t
Figure 5.10: Normalized concentration of C versus dimensionless time for the series reaction A → B → C for different values of k2/k1.
54 / 152 Applying the QSSA
Define component B to be a QSSA species. We write dc Bs = 0 = k c − k c dt 1 As 2 Bs That leads to k1 cBs = cAs (5.54) k2
Note cBs is not constant; it changes linearly with the concentration of A.
The issue is not if cB is a constant, but whether or not the B equilibrates quickly to its quasi-steady value.
55 / 152 Applying the QSSA
Substitution of cBs into the material balances for components A and C results in dc As = −k c (5.55) dt 1 As dc Cs = k c (5.56) dt 1 As The solutions to Equations 5.55 and 5.56 are presented in the following table for the initial condition cAs = cA0, cBs = cCs = 0.
56 / 152 Results of QSSA
k k Scheme I: A −→1 B −→2 C
Exact Solution QSSA Solution
−k1t −k1t cA = cA0e cAs = cA0e k k 1 −k1t −k2t 1 −k1t cB = cA0 e − e cBs = cA0 e k2 − k1 k2 1 −k t −k1t −k2t 1 cC = cA0 k2(1 − e ) − k1(1 − e (5.60)) cCs = cA0 1 − e (5.63) k2 − k1
Table: Exact and QSSA solutions for kinetic Scheme I.
57 / 152 Approximation error in C production
100
10−1
10−2
4 k2/k1=10 Cs
c −3 C 10 c − C
c 5
10 10−4
106 10−5
10−6 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
k1t
Figure 5.11: Fractional error in the QSSA concentration of C versus dimensionless time for the series reaction A → B → C for different values of k2/k1.
58 / 152 QSSA — Summary
The QSSA is a useful tool in reaction analysis. Material balances for batch and plug-flow reactors are ordinary differential equations. By setting production rates to zero for components that are QSSA species, their material balances become algebraic equations. These algebraic equations can be used to simplify the reaction expressions and reduce the number of equations that must be solved simultaneously. In addition, appropriate use of the QSSA can eliminate the need to know several difficult-to-measure rate constants. The required information can be reduced to ratios of certain rate constants, which can be more easily estimated from data.
59 / 152 Developing Rate Expressions from Complex Mechanisms
In this section we develop rate expressions for the mechanistic schemes presented in the introduction. The goal is to determine the rate in terms of measurable quantities, such as the concentrations of the reactants and products, and the temperature. The following procedure enables one to develop a reaction-rate expression from a set of elementary reactions.
60 / 152 Procedure
1 Identify the species that do not appear in equilibrium reactions and write a statement for the rate of production of these species using Equations 5.38 and 5.39.
Y −νij Y νij ri = ki cj − k−i cj (5.38) j∈Ri j∈Pi
n Xr Rj = νij ri (5.39) i=1 2 If the rate of production statement developed in step 1 contains the concentration of reaction intermediates, apply the QSSA to these species. 3 Perform the necessary algebraic manipulations and substitute the resulting intermediate concentrations into the rate of production statement from step 1.
61 / 152 Production rate of acetone
Example 5.2: Production rate of acetone The thermal decomposition of acetone is represented by the following stoichiometry
3CH3COCH3 −→ CO + 2CH4 + CH2CO + CH3COC2H5
2
62 / 152 Production rate of acetone
Use the following mechanism to determine the production rate of acetone [17]. You may assume the methyl, acetyl and acetone radicals are QSSA species.
k1 CH3COCH3 −→ CH3 + CH3CO
k2 CH3CO −→ CH3 + CO
k3 CH3 + CH3COCH3 −→ CH4 + CH3COCH2
k4 CH3COCH2 −→ CH2CO + CH3
k5 CH3 + CH3COCH2 −→ CH3COC2H5
63 / 152 Production rate of acetone
Solution Let the species be designated as: Species Formula Name A1 CH3COCH3 acetone A2 CH3 methyl radical A3 CH3CO acetyl radical A4 CO carbon monoxide A5 CH3COCH2 acetone radical A6 CH2CO ketene A7 CH4 methane A8 CH3COC2H5 methyl ethyl ketone
64 / 152 Production rate of acetone — QSSA
Write the production rate of acetone using Reactions 5.77 and 5.79
R1 = −k1c1 − k3c1c2 (5.82)
Species 2, 3 and 5 are QSSA species. Apply the QSSA to each of these species
R2 = 0 = k1c1 + k2c3 − k3c1c2 + k4c5 − k5c2c5 (5.83)
R3 = 0 = k1c1 − k2c3 (5.84)
R5 = 0 = k3c1c2 − k4c5 − k5c2c5 (5.85)
65 / 152 Production rate of acetone — QSSA
From Equation 5.84 k1 c3 = c1 (5.86) k2 Adding Equations 5.83, 5.84 and 5.85 gives
k1 c1 c5 = (5.87) k5 c2
66 / 152 Production rate of acetone—QSSA
Inserting the concentrations shown in Equations 5.86 and 5.87 into Equation 5.83 gives 2 k1k4 k3c2 − k1c2 − = 0 k5
s 2 k1 k1 k1k4 c2 = + 2 + (5.88) 2k3 4k3 k3k5 in which the positive sign is chosen to obtain a positive concentration. This result can be substituted into Equation 5.82 to give the rate in terms of measurable species. s 2 3 k1 k1k3k4 R1 = − k1 + + c1 (5.89) 2 4 k5
67 / 152 Production rate of acetone—QSSA
Equation 5.89 can be simplified to
R1 = −keff c1 (5.90)
by defining an effective rate constant s 2 3 k1 k1k3k4 keff = k1 + + 2 4 k5
68 / 152 Summary of acetone example
The production rate of acetone is first order in the acetone concentration. If, based on kinetic theories, we knew all the individual rate constants, we could calculate the rate of acetone production. Alternately, we can use Equation 5.90 as the basis for designing experiments to determine if the rate of production is first order in acetone and to determine the magnitude of the first-order rate constant, keff .
69 / 152 Nitric oxide example
Example 5.3: Production rate of oxygen Given the following mechanism for the overall stoichiometry
2NO2 + hν −→ 2NO + O2
derive an expression for the production rate of O2.
Assume atomic O and NO3 radicals are QSSA species. The production rate should be in terms of the reactant, NO2, and products, NO and O2. 2
70 / 152 Nitric oxide example
k1 NO2 + hν −→ NO + O
k2 O + NO2 )−*− NO3 k−2 k3 NO3 + NO2 −→ NO + O2 + NO2
k4 NO3 + NO −→ 2NO2
k5 O + NO2 −→ NO + O2
Let the species be designated as:
A1 = NO2, A2 = NO, A3 = O, A4 = NO3, A5 = O2
71 / 152 Nitric oxide example
The production rate of molecular oxygen is
R5 = k3c1c4 + k5c1c3 (5.97)
Species 3 and 4 are QSSA species. Applying the QSSA gives
R3 = 0 = k1c1 − k2c1c3 + k−2c4 − k5c1c3 (5.98)
R4 = 0 = k2c1c3 − k−2c4 − k3c1c4 − k4c2c4 (5.99)
Adding Equations 5.98 and 5.99
0 = k1c1 − k3c1c4 − k4c2c4 − k5c1c3 (5.100)
This result can be used to simplify the rate expression.
72 / 152 Nitric oxide example
Adding Equations 5.97 and 5.100
R5 = k1c1 − k4c2c4 (5.101)
The intermediate NO3 concentration is found by solving Equation 5.98 for c3.
k1 k−2 c4 c3 = + (5.102) k2 + k5 k2 + k5 c1 Substituting Equation 5.102 into Equation 5.99 and rearranging gives
k1k2c1 c4 = (5.103) (k2k3 + k3k5)c1 + (k2k4 + k4k5)c2 + k−2k5 The rate expression now can be found using Equations 5.101 and 5.103
k1k2k4c1c2 R5 = k1c1 − (5.104) (k2k3 + k3k5)c1 + (k2k4 + k4k5)c2 + k−2k5
73 / 152 Summary of nitric oxide example
Equation 5.104 is rather complex; but several terms in the denominator may be small leading to simpler power-law reaction-rate expressions. One cannot produce forms like Equation 5.104 from simple collision and probability arguments. When you see these forms of rate expressions, you conclude that there was a complex, multistep mechanism, which was reduced by the equilibrium assumption or QSSA.
74 / 152 Free-radical polymerization kinetics
Example 5.4: Free-radical polymerization kinetics Polymers are economically important and many chemical engineers are involved with some aspect of polymer manufacturing during their careers. Polymerization reactions raise interesting kinetic issues because of the long chains that are produced. Consider free-radical polymerization reaction kinetics as an illustrative example. A simple polymerization mechanism is represented by the following set of elementary reactions. 2
75 / 152 Free-radical polymerization reactions
Initiation:
k1 I −→ R1 Propagation:
kp1 R1 + M −→ R2 kp2 I is initiator R2 + M −→ R3
kp3 M is monomer R3 + M −→ R4 Mj is a dead polymer chain of length j . . Rj is a growing polymer chain of length j
kpj Rj + M −→ Rj+1 . . Termination:
ktmn Rm + Rn −→ Mm+n
76 / 152 Polymerization kinetics
Initiation. In free-radical polymerizations, the initiation reaction generates the free radicals, which initiate the polymerization reactions. An example is the thermal dissociation of benzoyl peroxide as initiator to form a benzyl radical that subsequently reacts with styrene monomer to form the first polymer chain. The termination reaction presented here is a termination of two growing polymer chains by a combination reaction.
77 / 152 Polymerization kinetics
Develop an expression for the rate of monomer consumption using the assumptions that kpj = kp for all j, and ktmn = kt is independent of the chain lengths of the two terminating chains. You may make the QSSA for all growing polymer radicals.
78 / 152 Solution
From the propagation reactions we can see the rate of monomer consumption is given by ∞ ∞ X X RM = − rpj = −kpcM cRj (5.105) j=1 j=1 The rate of the jth propagation reaction is given by
rpj = kpcM cRj
79 / 152 QSSA
Making the QSSA for all polymer radicals, we set their production rates to zero
P∞ rI − kpcR1 cM − kt cR1 j=1 cRj = 0 P∞ kpcR1 cM − kpcR2 cM − kt cR2 j=1 cRj = 0 P∞ kpcR2 cM − kpcR3 cM − kt cR3 j=1 cRj = 0 . . . + . . . (5.106) P∞ P∞ rI − kt i=1 cRi j=1 cRj = 0
80 / 152 Polymerization kinetics
We solve Equation 5.106 for the total growing polymer concentration and obtain
∞ X p cRj = rI /kt j=1 Substituting this result into Equation 5.105 yields the monomer consumption rate p RM = −kpcM rI /kt
From the initiation reaction, the initiation rate is given by rI = kI cI , which upon substitution gives the final monomer consumption rate
p √ RM = −kp kI /kt cI cM (5.107)
81 / 152 Polymerization summary
Consumption rate of monomer is first order in monomer concentration. We can therefore easily calculate the time required to reach a certain conversion. Notice that this result also provides a mechanistic justification for the production rate used in Chapter 4 in which monomer consumption rate was assumed linear in monomer concentration.
82 / 152 Ethane pyrolysis
Example 5.5: Ethane pyrolysis This example illustrates how to apply the QSSA to a flow reactor. We are interested in determining the effluent concentration from the reactor and in demonstrating the use of the QSSA to simplify the design calculations. Ethane pyrolysis to produce ethylene and hydrogen also generates methane as an unwanted reaction product. 2
83 / 152 Ethane pyrolysis
The following mechanism and kinetics have been proposed for ethane pyrolysis [10]
k1 C2H6 −→ 2CH3
k2 CH3 + C2H6 −→ CH4 + C2H5
k3 C2H5 −→ C2H4 + H
k4 H + C2H6 −→ H2 + C2H5
k5 H + C2H5 −→ C2H6
84 / 152 Rate constants
The rate constants are listed in the table −1 3 The preexponential factor A0 has units of s or cm /mol s for first- and second-order reactions, respectively.
Reaction A0 E(kJ/mol) 1 1.0 × 1017 356 2 2.0 × 1011 44 3 3.0 × 1014 165 4 3.4 × 1012 28 5 1.6 × 1013 0
Table 5.6: Ethane pyrolysis kinetics.
85 / 152 Reactor
The pyrolysis is performed in a 100 cm3 isothermal PFR, operating at a constant pressure of 1.0 atm and at 925 K. The feed consists of ethane in a steam diluent. The inlet partial pressure of ethane is 50 Torr and the partial pressure of steam is 710 Torr. The feed flowrate is 35 cm3/s.
86 / 152 Exact solution
The exact solution involves solving the set of eight species mass balances dN j = R (5.108) dV j subject to the feed conditions.
The results for C2H6, C2H4 and CH4 are plotted in the next figure.
The H2 concentration is not shown because it is almost equal to the C2H4 concentration.
Note the molar flowrate of CH4 is only 0.2% of the molar flowrate of the other products, C2H4 and H2. −6 The concentrations of the radicals, CH3, C2H5 and H, are on the order of 10 times the ethylene concentration.
87 / 152 Exact solution
3.5 × 10−5 6 × 10−8
3.0 × 10−5 methane 5 × 10−8
2.5 × 10−5 −8 ethylene 4 × 10 2.0 × 10−5 3 × 10−8 1.5 × 10−5
2 × 10−8 −5 molar flowrate (mol/s) 1.0 × 10 ethane 1 × 10−8 5.0 × 10−6
0.0 × 100 0 × 100 0 20 40 60 80 100 volume (cm3)
88 / 152 Find a simplified rate expression
Assuming the radicals CH3, C2H5 and H are QSSA species, develop an expression for the rate of ethylene formation. Verify that this approximation is valid.
89 / 152 Solution
The rate of ethylene formation is
RC2H4 = k3cC2H5 (5.109)
Next use the QSSA to relate C2H5 to stable reactants and products
RCH3 = 0 = 2k1cC2H6 − k2cCH3 cC2H6 (5.110)
RH = 0 = k3cC2H5 − k4cHcC2H6 − k5cHcC2H5 (5.111)
RC2H5 = 0 = k2cCH3 cC2H6 − k3cC2H5 + k4cC2H6 cH − k5cHcC2H5 (5.112) Adding Equations 5.110, 5.111 and 5.112 gives
k1 cC2H6 cH = (5.113) k5 cC2H5
90 / 152 Simplified rate expression
Inserting Equation 5.113 into Equation 5.111 yields 2 2 0 = k3k5cC2H5 − k4k1cC2H6 − k1k5cC2H6 cC2H5 s 2 k1 k1 k1k4 cC2H5 = + + cC2H6 (5.114) 2k3 2k3 k3k5 Finally s 2 k1 k1 k1k4 RC2H4 = k3 + + cC2H6 (5.115) 2k3 2k3 k3k5 This can be rewritten as
RC2H4 = kcC2H6 (5.116) in which s 2 k1 k1 k1k4 k = k3 + + 2k3 2k3 k3k5
At 925 K, k = 0.797 s−1. Let’s show that k1/k5 cH = (5.117) q 2 k1/2k3 + (k1/2k3) + k1k4/k3k5
2k1 cCH3 = (5.118)91 / 152 k2 Verification
The validity of the QSSA is established by solving the set of ODEs dN dN C2H6 = −r − r − r + r H2 = r dV 1 2 4 5 dV 4 dN dN CH4 = r H2O = 0 dV 2 dV dN C2H4 = r dV 3 subject to the initial molar flowrates. During the numerical solution the concentrations needed in the rate equations are computed using NC2H6 P cC2H6 = (5.119) NC2H6 + NCH4 + NC2H4 + NH2 + NH2O RT and Equations 5.114, 5.117 and 5.118.
Note we can neglect H, CH3 and C2H5 in the molar flowrate balance (Equation 5.119) because these components are present at levels on the order of −6 10 less than C2H6.
92 / 152 Verification
The next figure shows the error in the molar flowrate of ethylene that results from the QSSA. The error is less than 10% after 3.1 cm3 of reactor volume; i.e., it takes about 3% of the volume for the error to be less than 10%. Similarly, the error is less than 5% after 5.9 cm3 and less than 1% after 18.6 cm3. The figure shows that the QSSA is valid for sufficiently large reactors, such as 100 cm3. If the reactor volume were very small (a very short residence time), such as 20 cm3, the QSSA would not be appropriate.
93 / 152 Error in QSSA reduction
100
10−1
s 4 H 2 C 4 c
H −2
2 10 − C 4 c H 2 C c
10−3
10−4 0 10 20 30 40 50 60 70 80 90 100 Volume (cm3)
94 / 152 Is QSSA useful here?
What has been gained in this QSSA example? The full model solution requires the simultaneous solution of eight ODEs and the reduced model requires solving five ODEs. Both full and reduced models have five rate constants.
However, the results for the exact solution demonstrate that CH4 is a minor product at this temperature and this also would be found in the experimental data.
This suggests we can neglect CH4 in a species balance when computing mole fractions.
95 / 152 Further reduction
Therefore the mass action statement
k C2H6 −→ C2H4 + H2
does a reasonable job of accounting for the changes that occur in the PFR during ethane pyrolysis. QSSA analysis predicts the ethane pyrolysis rate should be first order in ethane concentration and the rate constant is k = 0.797 s−1. Using the reduced model
RC2H6 = −r RH2 = r we can solve the model analytically. The next figure shows that nearly equivalent predictions are made for the simplified scheme based on the mass action statement, Reaction 5.120 using Equation 5.116 for the rate expression.
96 / 152 Exact and reduced model
3.5 × 10−5
3.0 × 10−5
2.5 × 10−5 ethane ethylene 2.0 × 10−5
1.5 × 10−5
−5 molar flowrate (mol/s) 1.0 × 10
5.0 × 10−6
0.0 × 100 0 20 40 60 80 100 volume (cm3)
97 / 152 Summary of ethane pyrolysis example
This simple example demonstrates that when one has all the kinetic parameters the full model can be solved and the QSSA is not useful. The reverse situation of knowing a mechanism but not the rate constants could pose a difficult optimization problem when fitting all the rate constants to data. The QSSA analysis provides a framework for simplifying the kinetic expressions and removing difficult-to-estimate rate constants. In this example, one would expect a first-order expression to describe adequately the appearance of ethylene
98 / 152 Reactions at Surfaces
Our goal is to provide the student with a simple picture of surfaces and to provide a basis for evaluating heterogeneously catalyzed reaction kinetics and reaction-rate expressions. Keep in mind the goal is to develop a rate expression from a reaction mechanism. This rate expression can be used to fit model parameters to rate data for a heterogeneously catalyzed reaction. We present the elementary steps that comprise a surface reaction — adsorption, surface reaction and desorption. To streamline the presentation, the discussion is limited to gas-solid systems and a single chemical isotherm, the Langmuir isotherm. The interested student can find a more complete and thorough discussion of adsorption, surface reaction and desorption in numerous textbooks [5, 18, 1, 8, 11, 3].
99 / 152 Catalyst surface
A heterogeneously catalyzed reaction takes place at the surface of a catalyst. Catalysts, their properties, and the nature of catalytic surfaces are discussed in Chapter 7. For this discussion, we approximate the surface as a single crystal with a known surface order. The density of atoms at the low-index planes of transition metals is on the order of 1015 cm−2. Figure 5.16 presents the atomic arrangement of low-index surfaces for various metals.
100 / 152 Some common crystal structures of surfacesI
bcc (110)fcc(111) fcc (100)
bcc (100) fcc (110)
Figure 5.16: Surface and second-layer (dashed) atom arrangements for several low-index surfaces.
101 / 152 Some common crystal structures of surfacesII
This figure illustrates the packing arrangement and different combinations of nearest neighbors that can exist at a surface. An fcc(111) surface atom has six nearest-surface neighbors An fcc(100) surface atom has four nearest-surface neighbors An fcc(110) surface atom has two nearest-surface neighbors. The interaction between an adsorbing molecule and the surface of a single crystal depends on the surface structure.
102 / 152 CO oxidation example
Consider the oxidation of CO on Pd, an fcc metal. The following mechanism has been proposed for the oxidation reaction over Pd(111) [7]
O2 + 2S )−*− 2Oads
CO + S )−*− COads
COads + Oads −→ CO2 + 2S
in which the subscript ads refers to adsorbed species and S refers to vacant surface sites. This example illustrates the steps in a catalytic reaction — adsorption of reactants in Reactions 5.121 and 5.122, reaction of adsorbed components in Reaction 5.123, and desorption of products also shown as part of Reaction 5.123.
103 / 152 The adsorption process
Adsorption occurs when an incident atom or molecule sticks to the surface. The adsorbing species can be bound weakly to the surface (physical adsorption) or it can be held tightly to the surface (chemical adsorption). A number of criteria have been applied to distinguish between physical and chemical adsorption. In some cases the distinction is not clear. Table 5.7 lists several properties that can be used to distinguish between physisorption and chemisorption.
104 / 152 Physical and chemical adsorption
Property Chemisorption Physisorption amount limited to a monolayer multilayer possible specificity high none heat of adsorption typically > 10 kcal/mol low (2–5 kcal/mol) activated possibly generally not
Table 5.7: Chemisorption and physisorption properties.
The most distinguishing characteristics are the degree of coverage and the specificity.
105 / 152 Physical and chemical adsorption
Chemisorption: Chemical adsorption occurs when a chemical bond or a partial chemical bond is formed between the surface (adsorbent) and the adsorbate. The formation of a bond limits the coverage to at most one chemisorbed adsorbate for every surface atom. The upper limit need not have a one-to-one adsorbate to surface atom stoichiometry; for example, saturation can occur after one-third of all sites are occupied. Physisorption: Physical adsorption occurs once the partial pressure of the adsorbate is above its saturation vapor pressure. Physisorption is similar to a condensation process and has practically no dependence on the solid surface and the interaction between the adsorbate and adsorbent. Just as water vapor condenses on any cold surface placed in ambient, humid air, a gas such as N2 condenses on solids maintained at 77 K.
106 / 152 CO Adsorption
CO(g) CO(g)
kads kdes
O O O O O O C C C C C O C O C C O O O O C O O C O C C C C C
Pd (111) surface
Figure 5.17: Schematic representation of the adsorption/desorption process.
107 / 152 Langmuir model of adsorption
The Langmuir adsorption model assumes that all surface sites are the same. As long as the incident molecules possess the necessary energy, and entropy considerations are satisfied, adsorption proceeds if a vacant site is available. The Langmuir process also assumes adsorption is a completely random surface process with no adsorbate-adsorbate interactions. Since all sites are equivalent, and adsorbate-adsorbate interactions are neglected, the surface sites can be divided into two groups, vacant and occupied.
108 / 152 Langmuir model of adsorption
The surface-site balance is therefore Total number of Number of Number of surface sites = vacant sites + occupied sites per unit area per unit area per unit area
n Xs cm = cv + cj (5.124) j=1 Dividing by the total number of surface sites leads to the fractional form of the site balance. ns ns cv X cj X 1 = + = θv + θj (5.125) cm cm j=1 j=1
The units for cm often are number of sites per gram of catalyst.
109 / 152 Applying the Langmuir model
Consider again Figure 5.17 showing the model of a surface consisting of vacant sites and sites covered by chemisorbed CO. CO(g) CO(g)
kads kdes
O O O O O O C C C C C O C O C C O O O O C O O C O C C C C C
Pd (111) surface
110 / 152 Applying the Langmuir model
The CO in the gas phase is adsorbing on the surface and desorbing from the surface. The forward rate of adsorption and the reverse rate of desorption, are given by
k1 CO + S )−*− COads k−1
rads = k1cCOcv , rdes = k−1cCO The units of the rate are mol/time·area. The rate expressions follow directly from the reaction mechanism because the reactions are treated as elementary steps.
111 / 152 Adsorption at equilibrium
Equating the adsorption and desorption rates at equilibrium gives
k1cCOcv = k−1cCO
Solving for the surface concentration gives
k1 cCO = cCOcv = K1cCOcv (5.127) k−1 The conservation of sites gives
cv = cm − cCO (5.128)
Substituting Equation 5.128 into Equation 5.127 and rearranging leads to the Langmuir isotherm for single-component, associative adsorption
cmK1cCO cCO = (5.129) 1 + K1cCO
112 / 152 Langmuir isotherm
Dividing both sides of Equation 5.129 by the total concentration of surface sites cm leads to the fractional form of the single component, associative Langmuir adsorption isotherm. K1cCO θCO = (5.130) 1 + K1cCO
113 / 152 Langmuir isotherm
1 10.0
1.0
0.8 0.1
0.6
θCO
K=0.01 0.4
0.2
0 0 50 100 150 200 250
cCO (arbitrary units)
Figure 5.18: Fractional coverage versus adsorbate concentration for different values of the adsorption constant, K.
114 / 152 Figure 5.18 presents the general shape of the fractional coverage as a function of pressure for different values of K.
In the limit K1cCO 1, Equations 5.129 and 5.130 asymptotically approach cm and 1.0, respectively.
Since adsorption is exothermic, K1 decreases with increasing temperature and higher CO partial pressures are required to reach saturation. Figure 5.18 illustrates the effect of K on the coverage. In all cases the surface saturates with the adsorbate; however, the concentration at which this saturation occurs is a strong function of K.
Single component adsorption is used to determine the number of active sites, cm.
115 / 152 Fitting parameters to the Langmuir isotherm
The adsorption data can be fit to either the Langmuir isotherm (Equation 5.129) or to the “linear” form of the isotherm. The linear form is obtained by taking the inverse of Equation 5.129 and multiplying by cC0 c 1 c CO = + CO (5.131) cCO K1cm cm
A plot of the linear form provides 1/(K1cm) for the intercept and 1/cm for the slope.
116 / 152 CO data example
Example 5.6: Fitting Langmuir adsorption constants to CO data The active area of supported transition metals can be determined by adsorbing carbon monoxide. Carbon monoxide is known to adsorb associatively on ruthenium at 100◦C. Use the following uptake data for the adsorption of CO on 10 wt % Ru supported on ◦ Al2O3 at 100 C to determine the equilibrium adsorption constant and the number of adsorption sites.
PCO (Torr) 100 150 200 250 300 400 CO adsorbed (µmol/g cat) 1.28 1.63 1.77 1.94 2.06 2.21 2
117 / 152 Solution
The next two figures plot the data for Equations 5.129 and 5.131, respectively. From the data for the “linear” model, we can estimate using least squares the slope, 0.35 g/µmol, and the zero concentration intercept, 1.822 cm3/g, and then calculate
3 K1 = 0.190 cm /µmol
cm = 2.89 µmol/g
118 / 152 Original model
2.4 model data 2.2
2
1.8 mol/g) µ ( 1.6 CO c
1.4
1.2
1 4 6 8 10 12 14 16 18
3 cCO (µmol/cm )
Figure 5.19: Langmuir isotherm for CO uptake on Ru.
119 / 152 Linear model
8 model data 7 ) 3 6 (g/cm
CO 5 c / CO c 4
3
4 6 8 10 12 14 16 18
3 cCO (µmol/cm )
Figure 5.20: Linear form of Langmuir isotherm for CO uptake on Ru.
120 / 152 Multicomponent adsorption
We now consider multiple-component adsorption. If other components, such as B and D, are assumed to be in adsorption-desorption equilibrium along with CO, we add the following two reactions
k2 B + S )−*− Bads k−2 k3 D + S )−*− Dads k−3 Following the development that led to Equation 5.127, we can show the equilibrium surface concentrations for B and D are given by
cB = K2cB cv (5.132)
cD = K3cD cv
The site balance becomes
cv = cm − cCO − cB − cD (5.134)
121 / 152 Multicomponent adsorption
Substituting Equation 5.134 into Equation 5.127 and rearranging leads to a different isotherm expression for CO.
cmKCOcCO cCO = (5.135) 1 + K1cCO + K2cB + K3cD The terms in the denominator of Equation 5.135 have special significance and the magnitude of the equilibrium adsorption constant, when multiplied by the respective concentration, determines the component or components that occupy most of the sites.
Each of the terms Ki cj accounts for sites occupied by species j. From Chapter 3 we know the equilibrium constant is related to the exponential of the heat of adsorption, which is exothermic. Therefore, the more strongly a component is chemisorbed to the surface, the larger the equilibrium constant. Also, if components have very different heats of adsorption, the denominator may be well approximated using only one Ki cj term.
122 / 152 Dissociative adsorption
The CO oxidation example also contains a dissociative adsorption step
k1 O2 + 2S )−*− 2Oads k−1
Adsorption requires the O2 molecule to find a pair of adsorption sites and desorption requires two adsorbed O atoms to be adjacent. When a bimolecular surface reaction occurs, such as dissociative adsorption, associative desorption, or a bimolecular surface reaction, the rate in the forward direction depends on the probability of finding pairs of reaction centers. This probability, in turn, depends on whether or not the sites are randomly located, and whether or not the adsorbates are mobile on the surface.
123 / 152 Pairs of vacant sites
The probability of finding pairs of vacant sites can be seen by considering adsorption onto a checkerboard such as Figure 5.21.
A B
Figure 5.21: Two packing arrangements for a coverage of θ = 1/2.
Let two sites be adjacent if they share a common line segment (i.e., do not count sharing a vertex as being adjacent) and let θ equal the fraction of surface covered. The checkerboard has a total 24 adjacent site pairs, which can be found by counting the line segments on adjoining squares.
124 / 152 Pairs of vacant sites
Figure 5.21 presents two possibilities where half the sites (the shaded squares) are covered. For Figure 5.21A the probability an incident gas atom striking this surface in a random location hits a vacant site is 1 − θ and the probability a gas-phase molecule can dissociate on two adjacent vacant sites is p = 0. Figure 5.21B has 10 vacant adjacent site pairs. Therefore, the probability of dissociative adsorption is p = 10/24. The probability of finding vacant adsorption sites is 0 ≤ p ≤ 1/2 for these two examples because the probability depends on the packing arrangement.
125 / 152 Random packing
We next show for random or independent packing of the surface, the probability for finding pairs of vacant adjacent sites is p = (1 − θ)2. The probability for random packing can be developed by considering five different arrangements for a site and its four nearest neighbors, with the center site always vacant
12 3 4 5 The probability of any configuration is (1 − θ)nθm, in which n is the number of vacant sites and m is the number of occupied sites.
126 / 152 Random packing
Arrangement 1 can happen only one way. Arrangement 2 can happen four ways because the covered site can be in four different locations. Similarly, arrangement 3 can happen six different ways, 4 can happen four ways, and 5 can happen only one way. Therefore, the probability of two adjacent sites being vacant is
p1 + 4p2 + 6p3 + 4p4 + 0p5 (5.137)
127 / 152 Random packing
An incoming gas molecule can hit the surface in one of four positions, with one atom in the center and the other pointing either north, south, east or west. For random molecule orientations, the probabilities that a molecule hits the arrangement with a correct orientation are: p1, (3/4)4p2, (1/2)6p3 and (1/4)4p4. Therefore, the probability of a successful collision is
p = p1 + 3p2 + 3p3 + p4
128 / 152 Random packing
n m Use (1 − θ) θ for p1, p2, p3, p4
p = (1 − θ)5 + 3(1 − θ)4θ + 3(1 − θ)3θ2 + (1 − θ)2θ3 p = (1 − θ)2[(1 − θ)3 + 3(1 − θ)2θ + 3(1 − θ)θ2 + θ3]
Using a binomial expansion, (x + y)3 = x 3 + 3x 2y + 3xy 2 + y 3, the term in brackets can be written as
p = (1 − θ)2[(1 − θ) + θ]3 p = (1 − θ)2
129 / 152 Random packing — summary
Therefore, we use θv · θv to represent the probability of finding pairs of vacant sites for dissociative adsorption. The probability of finding other pair combinations is the product of the fractional coverage of the two types of sites.
130 / 152 Dissociative adsorption
Returning to Reaction 5.136
k1 O2 + 2S )−*− 2Oads k−1 We express the rates of dissociative adsorption and associative desorption of oxygen as
2 rads = k1cv cO2 (5.138) 2 rdes = k−1cO (5.139)
The units on k1 are such that rads has the units of mol/time·area. At equilibrium, the rate of adsorption equals the rate of desorption leading to p cO = K1cO2 cv (5.140) Combining Equation 5.140 with the site balance, Equation 5.124, and rearranging leads to the Langmuir form for single-component dissociative adsorption p cm K1cO2 cO = p (5.141) 1 + K1cO2
131 / 152 Langmuir isotherm for dissociative adsorption
The Langmuir isotherms represented in Equations 5.129 and 5.141 are different because each represents a different gas-surface reaction process. Both share the asymptotic approach to saturation at sufficiently large gas concentration, however.
132 / 152 Equilibrium CO and O surface concentrations
Determine the equilibrium CO and O surface concentrations when O2 and CO adsorb according to
k1 CO + S )−*− COads k−1 k2 O2 + 2S )−*− 2Oads k−2
133 / 152 Solution
At equilibrium adsorption rate equals desorption rate
r1 = 0 = k1cCOcv − k−1cCO 2 2 r2 = 0 = k2cO2 cv − k−2cO Solving for the surface coverages in terms of the concentration of vacant sites gives
cCO = K1cCOcv (5.142) p cO = K2cO2 cv (5.143)
The remaining unknown is cv , which can be found using the site balance
cm = cv + cCO + cO (5.144)
134 / 152 Solution
Combining Equations 5.142–5.144 yields
cm cv = p (5.145) 1 + K1cCO + K2cO2 We next substitute the vacant site concentration into Equations 5.142 and 5.143 to give the surface concentrations in terms of gas-phase concentrations and physical constants
cmK1cCO cCO = p 1 + K1cCO + K2cO2 p cm K2cO2 cO = p 1 + K1cCO + K2cO2
135 / 152 Production rate of CO2
Example 5.7: Production rate of CO2
Find the rate of CO2 production using the following reaction mechanism
k1 CO + S )−*− COads k−1 k2 O2 + 2S )−*− 2Oads k−2 k3 COads + Oads −→ CO2 + 2S
Assume the O2 and CO adsorption steps are at equilibrium.
Make a log-log plot of the production rate of CO2 versus gas-phase CO concentration at constant gas-phase O2 concentration. What are the slopes of the production rate at high and low CO concentrations? 2
136 / 152 Solution
The rate of CO2 production is given by
RCO2 = k3cCOcO (5.147) The surface concentrations for CO and O were determined in the preceding example,
cmK1cCO cCO = p 1 + K1cCO + K2cO2 p cm K2cO2 cO = p 1 + K1cCO + K2cO2
137 / 152 Solution
Substituting these concentrations onto Equation 5.147 gives the rate of CO2 production 2 p k3cmK1cCO K2cO2 RCO = (5.148) 2 p 2 1 + K1cCO + K2cO2
For ease of visualization, we define a dimensionless production rate ˜ 2 R = RCO2 /(k3cm)
and plot this production rate versus dimensionless concentrations of CO and O2 in the next figure.
138 / 152 Production rate of CO2
0.25 R˜ 0.2 0.15 0.1 0.05 0 100 10 1000 1 100 K c 10 0.1 1 CO 1 0.1 0.01 K2cO2 0.01
139 / 152 Production rate of CO2
Notice that the production rate goes to zero if either CO or O2 gas-phase concentration becomes large compared to the other. This feature is typical of competitive adsorption processes and second-order reactions. If one gas-phase concentration is large relative to the other, then that species saturates the surface and the other species surface concentration is small, which causes a small reaction rate. If we hold one of the gas-phase concentrations constant and vary the other, we are taking a slice through the surface of the figure. The next figure shows the result when each of the dimensionless gas-phase concentrations is held fixed at 1.0.
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100 R˜ versus K1cCO ˜ R versus K2cO2
10−1
10−2
R˜
10−3
10−4
10−5 10−3 10−2 10−1 100 101 102 103 104 dimensionless concentration
141 / 152 Production rate of CO2
Notice again that increasing the gas-phase concentration of either reactant increases the rate until a maximum is achieved, and then decreases the rate upon further increase in concentration. High gas-phase concentration of either reactant inhibits the reaction by crowding the other species off the surface.
Notice from the figure that log R˜ versus log(K1cC0) changes in slope from 1.0 to −1.0 as CO concentration increases.
As the concentration of O2 increases, the slope changes from 0.5 to −0.5. These values also can be deduced by taking the logarithm of Equation 5.148.
142 / 152 Hougen-Watson Kinetics
Rate expressions of the form of Equation 5.148 are known as Hougen-Watson or Langmuir-Hinshelwood kinetics [9]. This form of kinetic expression is often used to describe the species production rates for heterogeneously catalyzed reactions.
143 / 152 Summary
We introduced several concepts in this chapter that are the building blocks for reaction kinetics. Most reaction processes consist of more than one elementary reaction. The reaction process can be represented by a mass action statement, such as C2H6 −→ C2H4 + H2 in Example 5.5, but this statement does not describe how the reactants convert into the products. The atomistic description of chemical events is found in the elementary reactions. First principles calculations can be used to predict the order of these elementary reactions and the values of their rate constants.
144 / 152 Summary
The reaction order for elementary reactions is determined by the stoichiometry of the reaction Y −νij Y νij ri = ki cj − k−i cj j∈Ri j∈Pi Two assumptions are generally invoked when developing simplified kinetic rate expressions: 1 some of the elementary reactions are slow relative to the others; the fast reactions can be assumed to be at equilibrium. If only one step is slow, this single reaction determines all production rates, and is called the rate-limiting step; 2 the rate of formation of highly reactive intermediate species can be set to zero; the concentration of these intermediates is found from the resulting algebraic relations rather than differential equations.
145 / 152 Summary
Reactions at surfaces 1 The chemical steps involved in heterogeneous reactions (adsorption, desorption, surface reaction) are generally treated as elementary reactions. 2 In many cases, one reaction is the slow step and the remaining steps are at equilibrium. 3 For heterogeneous reactions, we add a site balance to account for the vacant and occupied sites that take part in the reaction steps.
146 / 152 NotationI
aj activity of species j cj concentration of species j cjs steady-state concentration of j cj concentration of species j on the catalyst surface cm total active surface concentration (monolayer coverage concentration) cv concentration of vacant surface sites fj fugacity of species j ◦ fj standard-state fugacity of pure species j Fj flux of species j from the gas phase to the catalyst surface gei degeneracy of the ith electronic energy level h Planck’s constant I moment of inertia for a rigid rotor Ii moment of inertia about an axis kB Boltzmann constant K reaction equilibrium constant m mass of a molecule (atom) Mj molecular weight of species j nj moles of species j Nj molar flowrate of species j p probability of finding adjacent pairs of sites P total pressure
147 / 152 NotationII
Pj partial pressure of species j Pi product species in reaction i q molecular partition function of species j V j qelec electronic partition function qrot rotational partition function qvib vibrational partition function Q partition function Qf volumetric flowrate at the reactor inlet ri reaction rate for ith reaction R gas constant Rj production rate for jth species Ri reactant species in reaction i t time T absolute temperature V (r) intermolecular potential energy VR reactor volume z compressibility factor i energy of the ith electronic level εi extent of the ith reaction θj fractional surface coverage of species j θv fractional surface coverage of vacant sites
148 / 152 NotationIII
νij stoichiometric number for the jth species in the ith reaction σ symmetry number, 1 for a heteronuclear molecule, 2 for homonuclear molecule τ lifetime of a component φj fugacity coefficient for species j
149 / 152 ReferencesI
A. W. Adamson. Physical Chemistry of Surfaces. John Wiley & Sons, New York, fifth edition, 1990.
M. Bodenstein. Eine Theorie der photochemischen Reaktionsgeschwindigkeiten. Zeit. physik. Chemie, 85:329–397, 1913.
M. Boudart and G. Djega-Mariadassou. Kinetics of Heterogenous Catalytic Reactions. Princeton University Press, Princeton, NJ, 1984.
D. L. Chapman and L. K. Underhill. The interaction of chlorine and hydrogen. The influence of mass. J. Chem. Soc. Trans., 103:496–508, 1913.
A. Clark. Theory of Adsorption and Catalysis. Academic Press, New York, 1970.
J. G. Ekerdt and A. T. Bell. Synthesis of hydrocarbons from CO and H2 over silica-supported Ru: Reaction rate measurements and infrared spectra of adsorbed species. J. Catal., 58:170, 1979.
T. Engel and G. Ertl. In D. P. Woodruff, editor, The Physics of Solid Surfaces and Heterogeneous Catalysis, volume four, page 73. Elsevier, Amsterdam, 1982.
150 / 152 ReferencesII
D. O. Hayward and B. M. W. Trapnell. Chemisorption. Butterworths, Washington, D.C., second edition, 1964.
O. A. Hougen and K. M. Watson. Solid catalysts and reaction rates. Ind. Eng. Chem., 35(5):529–541, 1943.
K. J. Laidler and B. W. Wojciechowski. Kinetics and mechanisms of the thermal decomposition of ethane. I. The uninhibited reaction. Proceedings of the Royal Society of London, Math and Physics, 260A(1300):91–102, 1961.
S. J. Lombardo and A. T. Bell. A review of theoretical models of adsorption, diffusion, desorption, and reaction of gases on metal surfaces. Surf. Sci. Rep., 13:1, 1991.
J. W. Moore and R. G. Pearson. Kinetics and Mechanism. John Wiley & Sons, New York, third edition, 1981.
Catalysis looks to the future. National Research Council, National Academy Press, Washington, D.C., 1992.
L. Pedersen and R. N. Porter. Modified semiempirical approach to the H3 potential-energy surface. J. Chem. Phys., 47(11):4751–4757, 1967.
J. C. Polanyi and J. L. Schreiber.
Distribution of reaction products (theory). Investigation of an ab initio energy-surface for F + H2 −→ HF + H. Chem. Phys. Lett., 29(3):319–322, 1974.
151 / 152 ReferencesIII
L. M. Raff, L. Stivers, R. N. Porter, D. L. Thompson, and L. B. Sims.
Semiempirical VB calculation of the (H2I2) interaction potential. J. Chem. Phys., 52(7):3449–3457, 1970.
F. O. Rice and K. F. Herzfeld. The thermal decomposition of organic compounds from the standpoint of free radicals. VI. The mechanism of some chain reactions. J. Am. Chem. Soc., 56:284–289, 1934.
G. A. Somorjai. Chemistry in Two Dimensions, Surfaces. Cornell University Press, Ithaca, New York, 1981.
R. C. Tolman. Duration of molecules in upper quantum states. J. Phys. Rev., 23:693, 1924.
T. Tur`anyi,A. S. Tomlin, and M. J. Pilling. On the error of the quasi-steady-state approximation. J. Phys. Chem., 97:163, 1993.
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